Uva1252 Twenty Questions

Twenty Questions

https://odzkskevi.qnssl.com/15b7eb4cd1f75f63cee3945b0b845e4f?v=1508411736

【题解】

dp[S1][S2]表示已经询问了S1，确定有S2，还需要至少多少次询问

dp[S1][S2] = max(min(dp[S1 | k][S2 | k], dp[S1 | k, S2 | j)) + 1

即枚举询问哪一个k，得到答复是有/没有中取还需要的询问次数最大的，然后

在所有的k中找最小的，表示到了S1，S2这个状态我要去询问k

预处理cnt[S1][S2]表示S2是S1的子集且属性为S2的物体的个数

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <algorithm>
 #include <sstream>
 #include <vector>
 #include <string>
 #include <cmath>
 #include <queue>
 #define min(a, b) ((a) < (b) ? (a) : (b))
 #define max(a, b) ((a) > (b) ? (a) : (b))
 
 inline void swap(int &a, int &b)
 {
     int tmp = a;a = b;b = tmp;
 }
 
 inline void read(int &x)
 {
     x = ;char ch = getchar(), c = ch;
     while(ch < '' || ch > '')c = ch, ch = getchar();
     while(ch <= '' && ch >= '')x = x *  + ch - '', ch = getchar();
     if(c == '-')x = -x;
 }
 
 const int INF = 0x3f3f3f3f;
 const int MAXN =  + ;
 const int MAXM =  + ;
 
 int ma,dp[ << MAXM][ << MAXM], cnt[ << MAXM][ << MAXM], n, m, num[MAXN];
 char s[MAXM];
 /*
 dp[s1][s2]表示已经询问了s1，确定有s2，还需要询问多少次
 */
 int DP(int s1, int s2)
 {
     if(cnt[s1][s2] <= )return ;
     if(dp[s1][s2] > -)return dp[s1][s2];
     dp[s1][s2] = INF;
     for(register int i = ;i < m;++ i)
     {
         if(s1 & ( << i))continue;
         dp[s1][s2] = min(dp[s1][s2], max(DP(s1 | ( << i), s2), DP(s1 | ( << i), s2 | ( << i))) + );
     }
     return dp[s1][s2];
 }
 
 int main()
 {
     while(scanf("%d %d", &m, &n) != EOF && m + n)
     {
         memset(num, , sizeof(num));
         memset(dp, -, sizeof(dp));
         memset(cnt, , sizeof(cnt));
         for(register int i = ;i <= n;++ i)
         {
             scanf("%s", s);
             for(register int j = ;j < m;++ j)
                 if(s[j] == '') num[i] |= ( << j);
         }
         ma =  << m;
         for(register int s1 = ;s1 < ma - ;++ s1)
             for(register int i = ;i <= n;++ i)
                 ++ cnt[s1][s1 & num[i]];
         printf("%d\n", DP(,));
     }
     return ;
 }

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